On level crossing in neutral integrodifferential equations

Applied Mathematics and Computation 119 (2001) 207±216 www.elsevier.com/locate/amc

On level crossing in neutral integrodi€erential equations A.A.S. Zaghrout *, E.A. Ali, Y.S. Ahmad Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr-City, P.O. Box 9019, Cairo 11765, Egypt

Abstract Necessary and sucient conditions are obtained for all solutions of neutral integrodi€erential equations of the form Z 1 d ‰x…t† ‡ px…t s† rx…t q†Š ‡ a k…s†x…t s† ds ˆ 0; t > 0 dt 0 to have at least one zero on … 1; 1†. Ó 2001 Elsevier Science Inc. All rights reserved. Keywords: Neutral integrodi€erential equation; Zero-crossings; Oscillations

1. Introduction Oscillation of delay di€erential equations has been extensively discussed by numerous authors, see [1,2,4,6,9]. However oscillatory behaviour of integrodi€erential equations especially those with in®nite delay has received little attention in the literature. In population dynamic models of delay di€erential equations, oscillations of model systems have certain ecological and evolutionary signi®cance, see [3,5,7]. For the equation of the form Z 1 d ‰x…t† cx…t s†Š ‡ a k…s†x…t s† ds ˆ 0; t > 0; dt 0

*

Corresponding author.

0096-3003/01/$ - see front matter Ó 2001 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 9 9 ) 0 0 2 5 8 - 1

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necessary and sucient conditions are obtained for all its solutions to have at least one zero, see [4] and [8]. The purpose of this paper is to derive a necessary and sucient condition for all solutions of neutral integrodi€erential equation of the form Z 1 d ‰x…t† ‡ px…t s† rx…t q†Š ‡ a k…s†x…t s† ds ˆ 0; t > 0 dt 0 …1:1† to have at least one zero on … 1; 1† (or to have zero crossings) where p, r, s, q and a are positive numbers and k : I ! I Z 1 k…s† ds ˆ 1; I ˆ ‰0; 1†: 0

Our condition is based on the nature of the roots of the characteristic equation associated with (1.1) which is k…1 ‡ pe

ks

re

kq

Z †‡a

1 0

ks

k…s†e

ds ˆ 0:

…1:2†

Indeed, we prove that every solution of (1.1) to have zero crossings i€ its characteristic equation (1.2) has no real roots under some conditions; on the other hand if there is a solution x or (1.1) such that either x…t† > 0 on … 1; 1† or x…t† < 0 on … 1; 1†, then such a solution is said to have no zero crossings. 2. Some useful lemmas In this section we establish some useful lemmas which will be used in proving our main theorem. Lemma 2.1. Suppose k : I ! I and k 6ˆ 0 on some subinterval of I. Let a, p, r 2 …0; 1†, 0 6 p < r < 1, s, q 2 I and s < q. If (1.2) has no real roots, then there exists a positive number m such that Z a

0

1

k…s†e

ks

ds P m ‡ k…1 ‡ pe

ks

re

kq

†;

k 2 R:

…2:1†

Proof. De®ne the function F as follows: F …k† ˆ k…1 ‡ pe

ks

with s < q, 0 6 p < r < 1.

re

kq

Z †‡a

1 0

k…s†e

ks

ds;

…2:2†

A.A.S. Zaghrout et al. / Appl. Math. Comput. 119 (2001) 207±216

Then we have Z F …0† ˆ a

1

0

F …k† ! 1

209

k…s† ds ˆ a > 0;

as k ! 1;

F …k† P k…1 ‡ …p

r†e

kq

Z †‡a

1

k…s†e

0

ks

ds ! 1

as k !

1;

and F …k† ˆ 0 has no real roots, a > 0. Hence inf F …k† > 0;

k 2 R:

…2:3†

If inf F …k† ˆ 0, then there exists a sequence fkn g, kn 2 R, jkn j < 1 such that k 2 R; F kn ! 0 as n ! 1. Since the sequence fkn g is bounded, there exists a convergent subsequence such that knk ! k and F …knk † ! 0

as nk ! 1:

Hence F is continuous in k, it follows that F …knk † ! F …k † ˆ 0

as nk ! 1

and hence k is a real root of F which is a contradiction. Thus there exists a positive number m such that F …k† ˆ k…1 ‡ pe

ks

re

kq

Z †‡a

1 0

k…s†e

ks

ds P m;

k 2 R:



Lemma 2.2. Let x…t† be an eventually positive solution of Eq. (1.1) and z0 …t† ˆ x…t† ‡ px…t

s†

rx…t

q†:

If conditions 0 6 p < r < 1, s < q hold, then the following assertions are valid: (i) z0 …t† is a monotone decreasing function, (ii) z0 …t† > 0 and (iii) limt!1 z0 …t† ˆ 0. Proof. Let x…t† is a positive solution of (1.1). Then we have, from (1.1), that z0 …t† ˆ x…t† ‡ px…t s† rx…t q† is decreasing as t increases. Thus if z0 …t† becomes zero or negative for some t ˆ t0 , then for all t > t0 we will have z0 …t† > 0; as a consequence there will exist d > 0 such that x…t† ‡ px…t

s†

rx…t

q† <

d

for t > t1 ˆ t0 ‡ 1:

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Hence x…t† <

d ‡ rx…t

< < ˆ

d ‡ …r

q†

px…t

s† <

d ‡ …r

p†‰ d ‡ …r

p†x…t

2q†Š




1

d ‡ …r r‡p

1

d ‡ x…t0 †e……t …r p†

p†n x…t

p†x…t

q†

nq†

t0 †=q†ln…r

p†

for t ˆ nq ‡ t0 ;

0 6 p < r < 1 and s < q: If t is large enough, then it will follow that   d x…t† 6 1 …r p† for large enough t, t 2 R which is impossible. Thus we conclude that z0 …t† ˆ x…t† ‡ px…t

s†

rx…t

q† > 0:

Set z1 …t† ˆ z0 …t† ‡ pz0 …t z2 …t† ˆ z1 …t† ‡ pz1 …t

s† s†

rz0 …t rz1 …t

q†; q†;

... Then zi …t†, i ˆ 0; 1; 2; . . . are solutions of (1.1). Furthermore z0 …t† 2 C 1 ‰t0 ; 1†, z1 …t† 2 C 2 ‰0; 1†, and they are eventually strictly monotone functions, we have Z 1 z_ 0 …t† ˆ ak…s†x…t s† ds < 0; …2:4† 0

Z z_ 1 …t† ˆ ‡

1

0

Z z1 …t† ˆ ‡

1

0

Z z_ 2 …t† ˆ

a

0

ak…s†z0 …t Z ak…s†

1

1

0

k…s†z1 …t

s† ds; ak…g†x…t s† ds;

…2:5† g† dg dn > 0; …2:6†

and so on. From (2.4) it follows that z0 …t† is positive and strictly decreasing and so lim z0 …t† ˆ ` 2 R:

t!1

…2:7†

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211

Then lim

t!1

d ‰z0 …t† ‡ pz0 …t dt

s†

ˆ lim z_ 1 …t† ˆ lim t!1

t!1

a

Z

rz0 …t 1 0

q†Š

k…s†z0 …t

s† ds ˆ

First we will prove that ` ˆ 0. For otherwise  1 lim ‰z0 …t† ‡ pz0 …t s† rz0 …t q†Š ˆ 1 t!1

a`:

if ` > 0; if ` < 0;

which contradicts (2.7). Thus we have established that thus limt!1 z0 …t† ˆ 0, which completes the proof.  3. Main result Our main result is the following: Theorem 3.1. Let a 2 …0; 1†; 0 6 p < r < 1; 0 6 s < q < 1 and that k is an eventually nonincreasing function. Suppose k 6ˆ 0 on some subinterval of I. A necessary and sufficient condition for nontrivial solution of (1.1) to have zero crossings is that the characteristic equation (1.1) has no real roots. Proof. The theorem will be proved in the contrapositive form there is nontrivial solutions of (1.1) to have zero crossings i€ the characteristic equation (1.2) has a real root. The necessity of the condition is easily seen, i.e. if (1.2) has a real root say l 2 R. Then (1.1) has a solution of the form x…t† ˆ Aelt ; A 2 R which has no zero crossings. Thus the necessity of the condition follows. For the suciently, that there is a non-oscillatory solution of (1.1) and, for the sake of contradiction, that (1.2) has no real roots. Thus, by Lemma 2.2, z0 …t† > 0 for t 2 R, also zi …t† > 0; i ˆ 1; 2; . . . for t 2 R. We note from Z 1 z_ 1 …t† ˆ a k…s†z0 …t s† ds; …3:1† 0

Z z_ 2 …t† ˆ

a

0

1

k…s†z1 …t

s† ds;

that

…3:2† Z

z_ 1 …t† ‡ p_z1 …t

s†

r_z1 …t

q† ‡ a

1 0

k…s†z1 …t

s† ds ˆ 0:

…3:3†

We can choose two positive numbers a; b (since k 6ˆ 0 on some interval of ‰0; 1Š), such that

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A.A.S. Zaghrout et al. / Appl. Math. Comput. 119 (2001) 207±216

Z Aˆ

a

b

k…s† ds > 0:

We have from (3.3)

Z

0 ˆ z_ 1 …t† ‡ p_z1 …t

s†

r_z1 …t Z

P z_ 1 …t† ‡ p_z1 …t† ‡ a

b a

q† ‡ a

k…s†z1 …t

1 0

k…s†z1 …t

s† ds

s† ds …since z1 is decreasing†;

i.e. z_ 1 …t† ‡

a 1‡p

Z

b a

k…s†z1 …t

s† ds 6 0:

Hence Z

b

z_ 1 …t† ‡

a z1 …t 1‡p

a†

z_ 1 …t† ‡

aA z1 …t 1‡p

a† 6 0:

a

k…s† ds 6 0;

or …3:4†

Integrating both sides of the inequality in (3.4) on ‰t …a=2†; tŠ we obtain Z t  a aA z1 …t† z1 t z1 …s a† ds 6 0; ‡ 2 1 ‡ p t …a=2† which leads to  z1 …t† z1 t This implies

aA a 

z1 t 1‡p 2

a aA z1 …t ‡ 2 1‡p  a 6 z1 t 2

a†
a 2

a 6 0: 2

 z1 …t† 6 z1 t

a : 2

A similar integration of the inequality in (3.4) on ‰t; t ‡ …a=2†Š leads to aA a  a

z1 t 6 z1 …t†: 1‡p 2 2 From (3.5) and (3.6) we conclude that " # 2 2 2…1 ‡ p† z1 …t a† 6 z1 …t†: aaA

…3:5†

…3:6†

…3:7†

A.A.S. Zaghrout et al. / Appl. Math. Comput. 119 (2001) 207±216

213

Consider now the set c…zi †; i ˆ 1; 2; . . . ; of real numbers de®ned by c…zi † ˆ fk P 0=_zi …t† ‡ kzi …t† 6 0 eventually for t > 0g:

…3:8†

For i ˆ 1, clearly k ˆ 0 2 c…z1 † and c…z1 † is nonempty; also c…z1 † is a subinterval of ‰0; 1†. The technique of our proof is to show that the existence of positive solution of (1.1) implies that the set c…z1 † has the following contradictory properties (i) and (ii), namely: (i) the set c…z1 † is bounded (ii) k 2 c…z1 † ) k ‡ m 2 c…z1 † where m is as in Lemma 2.1. To establish (i), we have to show the existence of an upper bound of c…z1 †. Integrating both sides of Z 1 z_ 1 …t† ‡ ak…s†z0 …t s† ds ˆ 0 on ‰t a; tŠ 0

we obtain z1 …t†

Z z1 …t

a† ‡ a

This implies that Z t Z 0 a z0 …s†k…u t a

a

t

Z

t a

1

0

Z s† ds ‡

u 0

z0 …s†k…u

Therefore for all large enough t > 0 Z 0 Z t a z0 …s†k…t s† ds ‡ z0 …s†k…t 1

1 6 z1 …t a

 s† ds du 6 z1 …t

a†:

 s† ds

s

a† 6

 s† ds du ˆ 0:

k…s†z0 …u

 2 1 2…1 ‡ p† z1 …t†; a aaA

which is the same as  2 Z 0 1 2…1 ‡ p† z0 …t s†k…s† ds 6 z1 …t† eventually: a a aaA 1

…3:9†

Now we have from (3.1) and (3.9)  2 Z 1 1 2…1 ‡ p† k…s†z0 …t s† ds < z_ l …t† ‡ z1 …t†; 0 ˆ z_ 1 …t† ‡ a a aaA 0 2

which shows that …1=a†‰…2…1 ‡ p††=…aaA†Š does not belong to c…z1 †, and thus the set c…z1 † is bounded. Also the same way for c…z2 †; c…z3 †; . . . ; from which 0 2 c…zi †; i ˆ 1; 2; . . . ; then the set c…zi † 6ˆ ; for any i ˆ 1; 2; . . . On the other hand, if k 2 c…zi †, then we can conclude that xi …t† ˆ 0…e kt † as t ! 1:

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In view of Henry's theorem which states that nonvanishing solution of Eq. (1.2) cannot tend to zero as t ! 1 faster than any exponential, we conclude that each set c…zi † is bounded.  Choose k 2 c…zi † such that k > sup c…zI † m and set /…t† ˆ ek t zI …t†: Relation (3.8) implies that eventually /0 …t† 6 0. From the inequality, the de®nition of the constant m and the fact that zI …t† is a solution of Eq. (1.2) it follows that eventually estimate (ii) holds. We de®ne 

k 2 c…zi †;

/1 …t† ˆ ek t zI …t†;

I ˆ 1; 2; . . . ;

…3:10†

then we have 

/01 …t† ˆ ek t ‰_z1 …t† ‡ k zI …t†Š 6 0;

…3:11†

showing that /1 …t† is nonincreasing; we have, from Lemma 2.1, (3.10) and (3.11) z_ 2 …t† ‡ …k ‡ m†z2 …t† Z 1 ˆ a k…s†z1 …t s† ds ‡ …k ‡ m†…z1 …t† ‡ pz1 …t 0 Z 1  ˆ a k…s†e k …t s† /1 …t s† ds 0

‡ …k ‡ m†‰e Z 1 ˆ a k…s†e 0

k t

/1 …t† ‡ pe

k …t s†

k …t s†

/1 …t

s†

re

s† ds ‡ …k ‡ m†e

/1 …t



k t

s†

rz1 …t

k …t q†

/1 …t

/1 …t†



‡ p…k ‡ m†e k …t s† /1 …t s† …k ‡ m†re k …t q† /1 …t  Z 1  k t k…s†e‡k s /1 …t† ds ‡ …k ‡ m†/1 …t† 6e a

q†

0



‡ …k ‡ m†e

‡k q

…pe

‡k s

/1 …t

s†

re

‡k q

/1 …t

…since s < q; 0 6 p < r < 1† Z 1   6e k t k…s†e‡k s /1 …t† dt ‡ …k ‡ m†/1 …t† a 0    ‡ …k ‡ m†e‡k q /1 …t q†…p r†  Z 1   ˆe kt k…s†e‡k s /1 …t† dt a

q††



0





‡ …k ‡ m†/1 …t†e‡k q …p

r† ‡ …k ‡ m†/1 …t†





q††

q†Š

A.A.S. Zaghrout et al. / Appl. Math. Comput. 119 (2001) 207±216

6e

k t



Z

/1 …t†

a

0

1



k…s†e‡k s ds ‡ k ‡ m



‡ …k ‡ m†pe 6e

k t



/1 …t†e‡k q …p

215

‡k s



r…k ‡ m†e

‡k q



r† 6 0:

Hence k ‡ m 2 c…z1 †. Similarly for i ˆ 2; 3; . . . ; we are led to a contradiction. Thus (1.2) cannot have a positive solution on … 1; 1†; similarly it cannot have a negative solution on … 1; 1†. This completes the proof.  Corollary 3.1. Assume that a; p; r; k; s; q are as in Theorem 3.1. If Z 1 Z 1 ea sk…s† ds > 1 r ‡ p or ea…r p ‡ 1† sk…s† ds > 1; 0

0

…3:12†

then all nontrivial solutions …1:2† to have at least one zero on … 1; 1†. Proof. Suppose that (1.2) has no zero solution on … 1; 1† by Theorem 3.1, the characteristic Eq. (1.2) has a real root, that is Z 1 F …0† ˆ k…1 ‡ pe ks re kq † ‡ a k…s†e ks ds ˆ 0 0

has a real root, since Z 1 F …k† ˆ a k…s† ds > 0; 0

F …k† > 0 for k P 0;

the real root has to be negative. Let k ˆ l; l > 0 be such a root; then l satis®es Z 1 a k…s†els ds ˆ l…1 ‡ pels relq † 6 l…1 ‡ pels rels †: 0

But l…1

r ‡ p† P l…1

which leads to …1

r ‡ p† P a

Z 0

1

…r

p†els † P a

sk…s†

els ds P ae ls

Z

1

0

Z 0

1

k…s†els ds;

sk…s† ds;

which contradicts the ®rst inequality of (3.12).

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A.A.S. Zaghrout et al. / Appl. Math. Comput. 119 (2001) 207±216

We also have

Z 1 Z 1 l > l…r p†els ‡ a k…s†els ds > l…r p† ‡ a k…s†els ds 0 0  Z 1  Z 1 ls k…s†e ds …r p† ‡ a k…s†els ds > a 0

0

leading to

Z

1 > a…r

p ‡ 1†

or

Z ae…r

p ‡ 1†

0

1

0

1

sk…s†

els ds > a…r ls

Z p ‡ 1†

1 0

sk…s†e ds

sk…s† ds < 1

and this contradicts the second inequality of (3.12). Thus the conclusion of the corollary follows.  References [1] O. Arino, I. Gyori, Necessary and sucient condition of neutral di€erential system with several delays, J. Di€erential Equations 81 (1989) 98±105. [2] K. Gopalsamy, Oscillatory properties of systems of ®rst order delay di€erential inequalities, Paci®c J. Math. 128 (1987) 299±305. [3] K. Gopalsamy, Harmless delays in a periodic system, J. Austral. Math. Soc. Ser. B 25 (1984) 349±365. [4] K. Gopalsamy, B.S. Lalli, Necessary and sucient conditions for zero crossing in integrodifferential equations, J. Tohoku. Math. 43 (2) (1991) 149±160. [5] M.K. Grammatikopoulos, Y.G. S®cas, I.P. Stavroulakis, Necessary and sucient conditions for oscillations of neutral equations with several coecients, J. Di€erential Equations 76 (1988) 294±311. [6] G.S. Ladde, V. Lakshmikantham, B.G. Zhang, Oscillation Theory of Di€erential Equations with Deviating Arguments, Marcel Dekker, New York, 1987. [7] R.K. Miller, Nonlinear Volterra Integral Equations, Benjamin, Menlo Park, 1971. [8] Y.G. S®cas, I.P. Stavroiclakis, Necessary and sucient conditions for oscillations of neutral di€erential equations, J. Math. Anal. Appl. 123 (1983) 494±507. [9] A.A.S. Zaghrout, S.H. Attalah, Asymptotic behaviour of neutral integrodi€erential equations, Acta Applicandae Mathematicae 42 (1996) 335±342.